Problem: Simplify the following expression: $y = \dfrac{-9x^2+43x+10}{x - 5}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(10)} &=& -90 \\ {a} + {b} &=& &=& {43} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-90$ and add them together. Remember, since $-90$ is negative, one of the factors must be negative. The factors that add up to ${43}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${45}$ $ \begin{eqnarray} {ab} &=& ({-2})({45}) &=& -90 \\ {a} + {b} &=& {-2} + {45} &=& 43 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 {-2}x) + ({45}x +{10}) $ Factor out the common factors: $ x(-9x - 2) - 5(-9x - 2)$ Now factor out $(-9x - 2)$ $ (-9x - 2)(x - 5)$ The original expression can therefore be written: $ \dfrac{(-9x - 2)(x - 5)}{x - 5}$ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ This leaves us with $-9x - 2; x \neq 5$.